Answer:
31.34 (to 2d.p.)
Step-by-step explanation:
Just use a calculator and type 42/134
-45.6 is the answer if you evaluate
A) We want to know when h=10 so
10=-16t^2+40t+3
16t^2-40t+7=0
Using the quadratic formula:
t=(40±√1152)/32
t≈0.189 and 2.311 seconds
So it is at 10 ft twice, about 2/10 of a second when it is rising and at 2 3/10 of a second when it is falling...
b.
The maximum height of the ball is when velocity is equal to zero, or dh/dt=0
dh/dt=-32t+40, dh/dt=0 when 32t=40, t=40/32=5/4=1.25 seconds
h(1.25)=28ft
So the maximum height of the ball occurs after 1.25 seconds and reaches a height of 28 feet.
